\(A=\frac{1}{1x3}+\frac{1}{3x5}+\frac{1}{5x7}+.......\frac{1}{13x15}=\frac{1}{2}x\frac{2}{1x3}+\frac{2}{3x5}.......+\frac{2}{13x15}\)

\(A=\frac{1}{2}x\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}\right)\)

Còn lại em nhân giống ở trên nhé

Đặt A = 1/15 + 1/35 + ... + 1/3135 

       A = 1/3.5 + 1/5.7 + ... + 1/55.57

     2A =  2/3.5 + 2/5.7 + ... + 2/55.57 

    2A = 1/3 - 1/5 + 1/5 - 1/7 + ... + 1/55 - 1/57 

    2A = 1/3 - 1/57 = 6/19 

      A = 3/19 

thanks các bn

Bài 1 :

\(A=3^0+3^1+3^2+3^3+...+3^{98}\)

\(A=\left(1+3+3^2\right)+.....+\left(3^{97}+3^{98}+3^{99}\right)\) ( Nhóm 3 số 1 nhé )

\(A=13+.....+3^{97}.13⋮13\left(\text{đ}pcm\right)\)

Bài 2 :

Theo ý a ta có : 

\(A=13+.....+3^{97}.13+3^{99}+3^{100}\)

\(A=13+.....+3^{97}.13+3^{99}.4⋮̸13\)

Bài 3 :

Để D chia hết cho 2 thì x chia hết cho 2

1. \(A=3^0+3^1+3^2+...+3^{98}\)

\(=\left(1+3+3^2\right)+\left(3^3+3^4+3^5\right)+...+\left(3^{96}+3^{97}+3^{98}\right)\)

\(=\left(1+3+3^2\right)+3^3\left(1+3+3^2\right)+...+3^{96}\left(1+3+3^2\right)\)

\(=13\left(1+3^3+...+3^{96}\right)\)chia hết cho \(13\).

2. \(B=3^0+3^1+3^2+3^3+...+3^{100}\)

\(=1+3+\left(3^2+3^3+3^4\right)+...+\left(3^{98}+3^{99}+3^{100}\right)\)

\(=4+3^2\left(1+3+3^2\right)+...+3^{98}\left(1+3+3^2\right)\)

\(=4+13\left(3^2+3^5+...+3^{98}\right)\)không chia hết cho \(13\).

3. \(D=\left(12.3+26.b+2022.c+x\right)\)chia hết cho \(2\)

\(\Leftrightarrow x⋮2\)(vì \(12.3⋮2,26b⋮2,2022c⋮2\))

Mình làm trước, chọn mình nha

Theo đề ta có a=5k+2

b=5q+3

13a+11b=13(5k+2)+11(5q+3)=65k+26+55q+33=(65k+55q)+59

Ta có 65k+55q chia hết cho 5 vì mỗi số hạng đều chia hết cho 5

59 chia 5 dư 4

Vậy 13a+11b chia 5 dư 4

a chia 5 dư 2=> a=5k+2       (k thuộc Z)

b chia 5 dư 3=> b=5h+3

=>(13.a+11.b)

=[13.(5k+2)+11.(5h+3)]

=65k+26+55h+33

=65k+55h+59

Vì 65k và 55h chia hết cho 5 mà 59 chia 5 dư 4

nên (13.a+11.b) chia cho 5 dư 4

a)\(\frac{4^5.9^4-2.6^9}{2^{10}.3^8+6^8.20}=\frac{\left(2^2\right)^5.\left(3^2\right)^4-2.\left(2.3\right)^9}{2^{10}.3^8+\left(3.2\right)^8.2^2.5}=\frac{2^{10}.3^8-2.2^9.3^9}{2^{10}.3^8+3^8.2^8.2^2.5}=\frac{2^{10}.3^8-2^{10}.3^9}{2^{10}.3^8+3^8.2^{10}.5}\)

\(=\frac{2^{10}.3^8.\left(1-3\right)}{2^{10}.3^8.\left(1+5\right)}=\frac{-2}{6}=\frac{-1}{3}\)

b) đặt A=2¹⁰⁰ - 2⁹⁹ + 2⁹⁸ - 2⁹⁷ +...+ 2² - 2

=>2A=2¹⁰¹-2¹⁰⁰+2⁹⁹-2⁹⁸+...+2³-2²

=>2A+A=2¹⁰¹-2¹⁰⁰+2⁹⁹-2⁹⁸+...+2³-2²+2¹⁰⁰ - 2⁹⁹ + 2⁹⁸ - 2⁹⁷ +...+ 2² - 2

=>3A=2¹⁰¹-2

=>A=\(\frac{2^{101}-2}{3}\)

thank kiu 

thank kiu

...............

Thank you bạn trieu dang

a) \(\frac{1}{3}+\frac{5}{6}:\left(x-2\frac{1}{5}\right)=\frac{3}{4}\)

=> \(\frac{1}{3}+\frac{5}{6}:\left(x-\frac{11}{5}\right)=\frac{3}{4}\)

=> \(\frac{5}{6}:\left(x-\frac{11}{5}\right)=\frac{3}{4}-\frac{1}{3}\)

=> \(\frac{5}{6}:\left(x-\frac{11}{5}\right)=\frac{5}{12}\)

=> \(x-\frac{11}{5}=\frac{5}{6}:\frac{5}{12}\)

=> \(x-\frac{11}{5}=2\)

=> \(x=2+\frac{11}{5}\)

=> \(x=\frac{21}{5}\)

thanks bn

1/15 +1/35 + 1/63 + ... + 1/3135

= 1/3*5 + 1/5*7 + 1/7*9 + ... + 1/55*57

= 1/3 - 1/5 + 1/5 - 1/7 + 1/7 - 1/9 + ... + 1/55 - 1/57

= 1/3 - 1/57

= 18/57

1/3 + 1/6 + 1/10 +  ... + 1/x(x + 1)*2 = 2009/2011

=> 2/6 + 2/12 + 2/20 + ... + 2/x(x + 1)*2 = 2009/2011

=> 2(1/6 + 1/12 + 1/20 + ... + 1/x(x + 1)) = 2009/2011

=> 2(1/2*3 + 1/3*4 + 1/4*5 + ... + 1/x(x + 1)) = 2009/2011

=> 2(1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 + ... + 1/x - 1/x + 1)  = 2009/2011

=> 2(1/2 - 1/x + 1) = 2009/2011

=> 1/2 - 1/x + 1 = 2009/4022

=> 1/x + 1 = 1/2011

=> x + 1 = 2011

=> x = 2010

vậy-

P = \(\frac{2}{3}.\frac{4}{5}.\frac{6}{7}...\frac{98}{99}\). CMR: P \(< \frac{1}{7}\)

Đề bài đây à

a, \(\dfrac{4}{7}\). \(\dfrac{a}{b}\) - \(\dfrac{1}{3}\) = \(\dfrac{1}{21}\)

     \(\dfrac{4}{7}\).\(\dfrac{a}{b}\)        =   \(\dfrac{1}{21}\) + \(\dfrac{1}{3}\)

     \(\dfrac{4}{7}\).\(\dfrac{a}{b}\)        = \(\dfrac{8}{21}\)

          \(\dfrac{a}{b}\)       = \(\dfrac{8}{21}\):\(\dfrac{4}{7}\)

           \(\dfrac{a}{b}\)      = \(\dfrac{2}{3}\)

b, \(\dfrac{a}{b}\) + \(\dfrac{2}{3}\).\(\dfrac{1}{3}\) = \(\dfrac{2}{3}\)

     \(\dfrac{a}{b}\) + \(\dfrac{2}{9}\) = \(\dfrac{2}{3}\)

      \(\dfrac{a}{b}\)        = \(\dfrac{2}{3}\) - \(\dfrac{2}{9}\)

       \(\dfrac{a}{b}\)       = \(\dfrac{4}{9}\)

c, \(\dfrac{a}{b}\) - \(\dfrac{1}{2}.\)\(\dfrac{2}{3}\) = \(\dfrac{2}{7}\)

    \(\dfrac{a}{b}\) - \(\dfrac{1}{3}\)     = \(\dfrac{2}{7}\)

    \(\dfrac{a}{b}\)           = \(\dfrac{2}{7}\) + \(\dfrac{1}{3}\)

    \(\dfrac{a}{b}\)          = \(\dfrac{13}{21}\)

d, \(\dfrac{11}{13}\): \(\dfrac{a}{b}\): \(\dfrac{2}{3}\) = 2\(\dfrac{7}{13}\)

     \(\dfrac{11}{13}\): \(\dfrac{a}{b}\):\(\dfrac{2}{3}\) = \(\dfrac{33}{13}\)

      \(\dfrac{11}{13}\): \(\dfrac{a}{b}\)    = \(\dfrac{33}{13}\) \(\times\) \(\dfrac{2}{3}\)

       \(\dfrac{11}{13}\): \(\dfrac{a}{b}\)   = \(\dfrac{66}{39}\)

                \(\dfrac{a}{b}\) = \(\dfrac{11}{13}\) : \(\dfrac{66}{39}\)

                 \(\dfrac{a}{b}\) = \(\dfrac{1}{2}\)

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